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# The EOQ Formula

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### The EOQ Formula

1. 1. The EOQ Inventory Formula James M. Cargal Mathematics Department Troy University – Montgomery Campus A basic problem for businesses and manufacturers is, when ordering supplies, to determinewhat quantity of a given item to order. A great deal of literature has dealt with this problem(unfortunately many of the best books on the subject are out of print). Many formulas andalgorithms have been created. Of these the simplest formula is the most used: The EOQ (economicorder quantity) or Lot Size formula. The EOQ formula has been independently discovered manytimes in the last eighty years. We will see that the EOQ formula is simplistic and uses severalunrealistic assumptions. This raises the question, which we will address: given that it is sounrealistic, why does the formula work so well? Indeed, despite the many more sophisticatedformulas and algorithms available, even large corporations use the EOQ formula. In general, largecorporations that use the EOQ formula do not want the public or competitors to know they usesomething so unsophisticated. Hence you might wonder how I can state that large corporations douse the EOQ formula. Let’s just say that I have good sources of information that I feel can be reliedupon.The Variables of the EOQ Problem Let us assume that we are interested in optimal inventory policies for widgets. The EOQformula uses four variables. They are: D: The demand for widgets in quantity per unit time. Demand can be thought of as a rate. Q: The order quantity. This is the variable we want to optimize. All the other variables are fixed quantities. C: The order cost. This is the flat fee charged for making any order and is independent of Q. h: Holding costs per widget per unit time. If we store x widgets for one unit of time, it costs us x@h. The EOQ problem can be summarized as determining the order quantity Q, that balances theorder cost C and the holding costs h to minimize total costs. The greater Q is, the less we will spendon orders, since we order less often. On the other hand, the greater Q is the more we spend oninventory. Note that the price of widgets is a variables that does not interest us. This is becausewe plan to meet the demand for widgets. Hence the value of Q has nothing to do with this quantity.If we put the price of widgets into our problem formulation, when we finally have finally solved theoptimal value for Q, it will not involve this term.
2. 2. The EOQ Inventory Formula by J. M. CargalThe Assumptions of the EOQ Model The underlying assumptions of the EOQ problem can be represented by Figure 1. The ideais that orders for widgets arrive instantly and all at once. Secondly, the demand for widgets isperfectly steady. Note that it is relatively easy to modify these assumptions; Hadley and Whitin[1963] cover many such cases. Despite the fact that many more elaborate models have beenconstructed for inventory problem the EOQ model is by far the most used.Figure 1 The EOQ ProcessAn Incorrect Solution Solving for the EOQ, that is the quantity that minimizes total costs, requires that weformulate what the costs are. The order period is the block of time between two consecutive orders.The length of the order period, which we will denote by P, is Q/D. For example, if the orderquantity is 20 widgets and the rate of demand is five widgets per day, then the order period is 20/5,or four days. Let Tp be the total costs per order period. By definition, the order cost per order periodwill be C. During the order period the inventory will go steadily from Q, the order amount, to zero.Hence the average inventory is Q/2 and the inventory costs per period is the average cost, Q/2, timesthe length of the period, Q/D. Hence the total cost per period is: Q Q Q2 h TP  C  h  C 2 D 2DIf we take the derivative of Tp with respect to Q and set it to zero, we get Q = 0. The problem issolved by the device of not ordering anything. This indeed minimizes the inventory costs but at thesmall inconvenience of not meeting demand and therefore going out of business. This is what manypeople, perhaps most people do, when trying to solve for the EOQ the first time.The Classic EOQ Derivation The first step to solving the EOQ problem is to correctly state the inventory costs formula.This can be done by taking the cost per period Tp and dividing by the length of the period, Q/D, toget the total cost per unit time, Tu: CD Qh Tu   Q 2 2
3. 3. The EOQ Inventory Formula by J. M. CargalIn this formula the order cost per unit time is CD/Q and Qh/2 is the average inventory cost per unittime. If we take the derivative of Tu with respect to Q and set that equal to 0, we can solve for theeconomic order quantity (where the exponent * implies that this is the optimal order quantity): 2CD Q*  hIf we plug Q* into the formula for Tu, we get the optimal cost, per unit time: Tu  * 2CDhThe Algebraic Solution It never hurts to solve a problem in two different ways. Usually each solution technique willyield its own insights. In any case, getting the same answer by two different methods, is a great wayto verify the result. If we multiply the formula for Tu on both sides by Q, we get, after a littlerearranging, a quadratic equation in Q: Q2 h  QTu  CD  0 2Next we divide through by h/2 to change our leading coefficient to 1. Completing the square, weget: 2  Tu  2CD Tu2 Q    2  0  h h hNote that this equation has two variables. Tu is a function of Q (and could well be written as Tu(Q)).Hence we have a curve in the Cartesian plane with axes labeled Q and Tu. We want the value of Q 2CDthat minimizes Tu. Notice that the term is a constant. Writing that constant as k, and hrearranging the equation, we get: 2 Tu2  T  Q u  k h2  h 3
4. 4. The EOQ Inventory Formula by J. M. CargalIf we set the quadratic term to zero, then Tu  h k . Any change in the quadratic term from zeroincreases the size of Tu. Hence the optimal size of Tu is h k which just happens to be Tu 2CDh which is the value we found earlier. The quadratic term is zero if and only if Q  . hThis gives us the identity Q*h  Tu* .An Example It is useful at this point to consider a numerical example. The demand for klabitz’s is 50 perweek. The order cost is \$30 (regardless of the size of the order), and the holding cost is \$6 perklabitz per week. Plugging these figures into the EOQ formula we get: 2  30  50 Q*   22.36 6This brings up a little mentioned drawback of the EOQ formula. The EOQ formula is not an integerformula. It would be more appropriate if we ordered klabitz’s by the gallon. Most of the time, thenearest integer will be the optimal integer amount. In this case, the total inventory cost Tu is \$134.18per week when we order 22 klabitz’s. If instead, we order 23 klabitz’s the cost is \$134.22. 1600 1400 1200 Total Cost 1000 800 600 Min 400 200 0 1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 Order Quantity Figure 2 Total Cost as a Function of Order Quantity 4
5. 5. The EOQ Inventory Formula by J. M. Cargal 1600 1400 1200 Holding Cost Order Cost 1000 800 600 Min 400 200 0 1 6 1116212631364146515661667176 Order Quantity Figure 3 Order and Holding Costs A graph of this problem is illuminating: Figure 2. Because the graph is so flat at the optimalpoint, there is very little penalty if we order a slightly sub-optimal quantity. We can betterunderstand the graph if we view the combined graphs of the order costs and the holding costs givenin Figure 3. The basic shapes of all three graphs (total costs, order costs, holding costs) are alwaysthe same. The graph of order costs is a hyperbola; the graph of holding costs is linear; and as a dTuresult the graph of the total costs (Tu) is convex. This can also be seen in that the function is dQincreasing and the function d 2 T 2CD  dQ2 Q3is positive every where. If we plug the optimal quantity, Q*, into this last formula we get: d 2T * 2hCD  dQ2 4h2CD 5